Question

the triple points of near and carbon dioxide are 24.5 7 Kelvin and 216.55 Kelvin respectively .Express the temperatures on the Celsius and Fahrenheit scales ​

Answer 1

Explanation:

$1) \: \frac{K - 273}{100} = \frac{C - 0}{100 } \\ 24.57 - 273 = C \\ C = - 248 .43° \\ \\ \\ \frac{K - 273}{100} = \frac{F - 32}{180} \\ \frac{24.57 - 273}{5} = \frac{F - 32}{9} \\ 9( - 248.43) = 5F - 160 \\ - 2235.87 + 160 = 5F \\ \frac{ - 2075.85}{5 } = F \\ F = - 415.17°\\ \\ 2) \frac{K - 273}{100} = \frac{C - 0}{100} \\ 216.55 - 273 = C \\ C = - 56.45 °\\ \\ \frac{K - 273}{100} = \frac{F - 32}{180} \\ \frac{216.55 - 273}{5} = \frac{F - 32}{9} \\ 9( - 56.45) = 5F - 160 \\ - 508. 05 + 160 = 5F \\ F = \frac{ - 348.05}{5} \\ F = - 69.61°$.

Answer 2

Explanation:

concept :-

K = C + 273.15

and C/5 = ( F - 32)/9

Where, K is Kelvin , C is Calcius and F is Fahrenheit.

For 24.57 K

In calcius ,

C = 24.57 - 273.15

C = -248.58° C

In Fahrenheit ,

-248.58/5 = ( F - 32)/9

F = -248.58 × 9/5 + 32

= -414.44°F

For 216.55 K,

in Celsius ,

C = 216.55 - 273.15

= -56.6°C

In Fahrenheit,

-56.6/5 = ( F - 32)/9

F = -56.6×9/5 + 32

= -69.88°F